A kettle boils 0.6kg of water. After some time the temperature of the water has decreased to 83 degrees. The specific heat capacity of water is 4200J/kg. Calculate the energy transferred to the surroundings.
The variables we are given are mass (m), time (t), temperate CHANGE (delta T) and SHC (c) and we are asked for energy. Think of the equations for energy we know: mgh, 1/2mv2, mcdeltaT...the first two energy equations are not relevant to this situationThe equation needed is Q = m x c x deltaTdelta T is the change in temperature. Water cools from boiling therefore: 100-deltaT = 83----> delta T = 17Q = 0.6 x 4200 x 17 = 42840 J
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